Regular expression of two digit number where two digits are not same - regex

I am trying to write a regular expression that will match a two digit number where the two digits are not same.
I have used the following expression:
^([0-9])(?!\1)$
However, both the strings "11" and "12" are not matching. I thought "12" would match. Can anyone please tell me where I am going wrong?

You need to allow matching 2 digits. Your regex ^([0-9])(?!\1)$ only allows 1 digit string. Note that a lookahead does not consume characters, it only checks for presence or absence of something after the current position.
Use
^(\d)(?!\1)\d$
^^
See demo
Explanation of the pattern:
^ - start of string
(\d) - match and capture into Group #1 a digit
(?!\1) - make sure the next character is not the same digit as in Group 1
\d - one digit
$ - end of string.

Related

Regular Expression to Validate Monaco Number Plates

I would like to have an expression to validate the plates of monaco.
They are written as follows:
A123
123A
1234
I started by doing:
^[a-zA-Z0-9]{1}?[0-9]{2}?[a-zA-Z0-9]{1}$
But the case A12A which is false is possible with that.
You can use
^(?!(?:\d*[a-zA-Z]){2})[a-zA-Z\d]{4}$
See the regex demo. Details:
^ - start of string
(?!(?:\d*[a-zA-Z]){2}) - a negative lookahead that fails the match if there are two occurrences of any zero or more digits followed with two ASCII letters immediately to the right of the current location
[a-zA-Z\d]{4} - four alphanumeric chars
$ - end of string.
You can write the pattern using 3 alternatives specifying all the allowed variations for the example data:
^(?:[a-zA-Z][0-9]{3}|[0-9]{3}[a-zA-Z]|[0-9]{4})$
See a regex demo.
Note that you can omit {1} and
To not match 2 chars A-Z you can write the alternation as:
^(?:[a-zA-Z]\d{3}|\d{3}[a-zA-Z\d]|\d[a-zA-Z\d][a-zA-Z\d]\d)$
See another regex demo.
So it needs 3 connected digits and 1 letter or digit.
Then you can use this pattern :
^(?=.?[0-9]{3})[A-Za-z0-9]{4}$
The lookahead (?=.?[0-9]{3}) asserts the 3 connected digits.
Test on Regex101 here

Regex to block more than 3 numbers in a string

I am trying to block any strings that contain more than 3 numbers and prevent special characters. I have the special characters part down. I'm just missing the number part.
For example:
"Hello 1234" - Not Allowed
"Hello 123" - Allowed
I've tried the following:
/^[!?., A-Za-z0-9]+$/
/((^[!?., A-Za-z]\d)([0-9]{3}+$))/
/^((\d){2}[a-zA-Z0-9,.!? ])*$/
The last one is the closest I got as it prevents any special characters and any numbers from being entered at all.
I've looked through previous posts, but am coming up short.
Edit for clarification
Essentially I'm trying to find a way to prevent customers from entering PII on a form. No submission should be allowed that contains more than 3 numbers in a string.
Hello1234 - Not allowed
12345 - Not allowed
1111 - not allowed
No where in the comment section when the user enters the string should there be more than 3 numbers in total.
About the patterns that you tried
^[!?., A-Za-z0-9]+$ The pattern matches 1+ times any of the listed, including 1 or more digits
((^[!?., A-Za-z]\d)([0-9]{3}+$)) If {3}+ is supported, the pattern matches a single char from the character class, 1 digit followed by 3 digits
^((\d){2}[a-zA-Z0-9,.!? ])*$ The pattern repeats 0+ times matching 2 digits and 1 of the listed in the character class
You can use a negative lookahead if that is supported to assert not 4 digits in a row.
^(?!.*\d{4})[a-zA-Z0-9,.!? ]+$
regex demo
If there can not be 4 digits in total, but 0-3 occurrences:
^[a-zA-Z,.!? ]*(?:\d[a-zA-Z,.!? ]*){0,3}$
Explanation
^ Start of string
[a-zA-Z,.!? ]* Match 0+ times any of the listed (without a digit)
(?:\d[a-zA-Z,.!? ]*){0,3} Repeat 0 - 3 times matching a single digit followed by optional listed chars (Again without a digit)
$ End of string
regex demo
If you don't want to match an empty string and a lookahead is supported:
^(?!$)[a-zA-Z,.!? ]*(?:\d[a-zA-Z,.!? ]*){0,3}$
See another regex demo
Here is my two cents:
^(?!(.*\d){4})[A-Za-z ,.!?\d]+$
See the online demo
^ - Start string anchor.
(?! - Open a negative lookahead.
( - Open capture group.
.*\d - Match anything other than newline up to a digit.
){4} - Close capture group and match it 4 times.
) - Close negative lookahead.
[A-Za-z ,.!?\d]+ - 1+ Characters from specified class.
$ - End string anchor.
I think it should cover what you described.
Assuming you mean <= 3 digits, this may be a naive one but how about
[ALLOWED_CHARS]*[0-9]?[ALLOWED_CHARS]*[0-9]?[ALLOWED_CHARS]*[0-9][ALLOWED_CHARS]*?
Fill [ALLOWED_CHARS] to whatever you define is not special character and nums.

How to validate pattern AB12CD

The requirement is "each 2 digits must be only numbers or only text" - so valid patterns are AB-12-CD or 12-AB-CD or AB-CD-12, 12-34-AB.
The below suggested pattern working fine without hyphen but if we want to add a hyphen in between, how to do?
\b(?=[A-Z\d][A-Z])(?=[A-Z\d]\d)(?:[A-Z]{2}|\d{2})+\b
Repeat 1 or more times matching either 2 uppercase chars or 2 digits. Note that there are no hyphens present in the example data.
\b(?:[A-Z]{2}|\d{2})+\b
Regex demo
If there must be a digit and an uppercase character present, you could also use a positive lookahead:
\b(?=[A-Z\d]*[A-Z])(?=[A-Z\d]*\d)(?:[A-Z]{2}|\d{2})+\b
Regex demo

How to match a 10 digit phone number which may or maynot be have a 2 or 3 digit country code.The country code is not to be matched

Example string
fgcfghhfghfgch1234567890fghfghfgh fhghghfgh+916546546165fghfghfghfgh fhfghfghfghfgh+915869327425ghfghfghfgh
I want to match
1234567890
6546546165
5869327425
In essence i would like to do something like this (?<=\+\d{2})?\d{10}.
Match 10 digits \d{10} which may follow ? a country code in format: \+\d{2}.
What would be a correct regular expression to do this?
Also,
What to do if the country code could possibly be even 3 digit long.
e.g.
+917458963214
+0047854123698
match 7854123698 and 7458963214.
Your expected matches all appear to be immediately followed with a char other than a digit.
I suggest making the pattern inside the positive lookbehind optional and adding (?!\d) lookahead at the end to fail the match (and thus triggering backtracking in the lookbehind) if the ten digits are immediately followed with a digit:
(?<=(?:\+\d{2,3})?)\d{10}(?!\d)
See the regex demo. Details:
(?<=(?:\+\d{2,3})?) - a positive lookbehind that requires + and two or three digits or an empty string immediately to the left of the current location
\d{10} - ten digits
(?!\d) - no digit allowed immediately on the right.
However, as in 99% you can access captured substrings, you should just utilize a capturing group:
(?:\+\d{2,3})?(\d{10})
See this regex demo. Your values are in Group 1.

Match against 1 hyphen per any number of digit groups

I'm trying to come up with some regex to match against 1 hyphen per any number of digit groups. No characters ([a-z][A-Z]).
123-356-129811231235123-1235612346123451235
/[^\d-]/g
The one above will match the string below, but it will let the following go through:
1223--1235---123123-------
I was looking at the following post How to match hyphens with Regular Expression? for an answer, but I didn't find anything close.
#Konrad Rudolph gave a good example.
Regular expression to match 7-12 digits; may contain space or hyphen
This tool is useful for me http://www.gskinner.com/RegExr/
Assuming it can't ever start with a hyphen:
^\d(-\d|\d)*$
broken down:
^ # match beginning of line
\d # match single digit
(-\d|\d)+ # match hyphen & digit or just a digit (0 or more times)
$ # match end of line
That makes every hyphen have to have a digit immediately following it. Keep in mind though, that the following are examples of legal patterns:
213-123-12314-234234
1-2-3-4-5-6-7
12234234234
gskinner example
Alternatively:
^(\d+-)+(\d+)$
So it's one or more group(s) of digits followed by hyphen + final group of digits.
Nothing very fancy, but in my tests it matched only when there were hyphen(s) with digits on both sides.